Saturday, December 16, 2006

 

Algebraic integers is a flawed ring

Excerpts taken from the math blog with some editing for size and clarity.

Simple algebra shows ring of algebraic integers is critically flawed by revealing factors blocked by special definition of ring. Significant error which has to be addressed by the mathematical community but it is not clear who has the nerve or the will to do anything about it.

Start with

175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

with functions f(x) and g(x), and an integral domain with more to be specified later.

Choose to multiply through f(x) + 1 by 2 and multiply both sides by 7:

7*(175x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)

note that the above is equivalent to

(49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (2f(x) + 2)*(7g(x) + 7)

where 7 has been distributed through g(x) + 1.

Now introduce additional functions a_1(x) and a_2(x), and let

2f(x) = 5a_1(x) + 5

and

7g(x) = 5a_2(x)

make the substitutions:

(49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (5a_1(x) + 7)*(5a_2(x) + 7)

solve for the a's using

a_1(x)*a_2(x) = 49x^2 - 14x

and

a_1(x) + a_2(x) = 7x-1

so it follows they are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

And that second quadratic gives the critical result.

For instance, let x=1, then

a^2 - 6a + 35 = 0

and

7g(1) = 5a_2(1)

would seem to indicate that a_2(1) has 7 as a factor, but so far it is only specified (as I added to the original argument from the blog above) that the domain is integral. But go to the ring of algebraic integers and it can be shown that 7 is not a factor of a_2(x) in that ring.

That is the back breaker result. The arbitrariness of the selection of g(x) to be multiplied by 7, completes the simple proof as you could have gone the other way:

175x^2 - 15x + 2 = (f(x) + 1)*(2g(x) + 2)

to get the same equation for the a's, but now with

7f(x) = 5a_2(x)

so now f(x) would also need to divide off factors from 7. QED

The argument with evens and odds is remarkably succinct in explaining the result.

What follows is just taken from the blog with very limited editing:

Well think if you wish to have only evens. Now consider that 2 is now not a factor of 6 as that will give you the odd 3, so just like that, with an arbitrary rule that excludes some numbers you get a result counter to what you know, as 2 is a factor of 6.

With algebraic integers, the rule is that an algebraic integer must be the root of some monic polynomial with integer coefficients, and the algebra above proves that some numbers are excluded by that rule so you get the odd result.

But now let's solve for the a's with x=1, as using the quadratic formula with

a^2 - 6a + 35 = 0

gives

a = 3 +/- sqrt(-26)

and you may wonder, which root has 7 as a factor?

The answer is, one root but it's ambiguous as to which as the sqrt() gives two answers while many of you may think that you can say that 3 + sqrt(-26) is one number while 3 - sqrt(-26) is another, but consider

1 + sqrt(4)

and if you say that's 3, well, what about -2? In the mathematical world I live in

(-2)(-2) = 4

so that is two solutions which can be represented as

1 +/- 2

and this may seem like a boring exercise given the importance of what I have above, but a lot of people are taught wrong on this point, and it's worth pointing out that when you have sqrt(), you have two numbers, no matter what.

Some math people will say that, no, they have one number as they have DEFINED the square root to be only the positive and they will also claim they can find a factor of 3 + sqrt(-26) in the ring of algebraic integers, so I'm wrong and there is nothing wrong with what they do.

But check this out:

(1 +/- 2)(3 +/- 2) = (7 +/- 8)

so you can do algebra with functions that have two values and even find factors that way, without it changing a thing.

(An interesting way to show the flaw in some objections I've noted from posters who seem to be taken seriously—though I don't know why—by readers on these newsgroups.)

Oddly enough though mathematicians got it wrong!

Some time ago some of them decided they didn't like functions that gave more than one answer so they defined sqrt() to have only one solution. But that couldn't get rid of the other solution! So -2 is still a solution to sqrt(4) as (-2)(-2) = 4, but math students learn to think it's not a solution to the square root function because it is defined not to be. That weird little error has consequences, as you have generations of people who think that 1+sqrt(2) is a single number different from 1-sqrt(2) when both actually give the same two numbers—just in different order.

Like

1+/-2 versus 1-/+2

as the first way gives 3 and -1, while the second gives -1 and 3.

So it's not hard to understand what is happening mathematically, and easy to prove a problem with the ring of algebraic integers, while also pointing out a weird thinking error with square roots which currently plagues a lot of people.

Want a paper? Then go to my page on this at my Extreme Mathematics group:

http://groups-beta.google.com/group/extrememathematics/web/non-polynomial-factorization-paper





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